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Likurg_2 [28]
3 years ago
5

What is the area of the triangle?

Mathematics
2 answers:
garri49 [273]3 years ago
7 0
40 right? Because base times height divided by two is area of a triangle
yawa3891 [41]3 years ago
3 0

Answer:

52

Step-by-step explanation:

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Marlon earned $8.50 per hour plus an additional $80 in tips waiting tables on Saturday. He earned at least $131 in all. If h rep
ella [17]

Answer:

<h2>8.50h + 80 ≥131</h2>

Step-by-step explanation:

The answer is not in the option

Step one:

given data:

we are told that the hourly earning= $8.50

then additional tip=$80

total earnings=$131.

Step two:

the linear function for the total earning is the same as the equation of a line

y=mx+c

where

y represents the total earning of $131

m represents the hourly earning= $8.50

x represents the number of hours h

c represents the tip of $80

The expression for the situation is modeled as

<h2>8.50h + 80 ≥131</h2>
8 0
3 years ago
Brook rode her bike 7 miles in 30 minutes. Joshua rode his bike 25 miles in 2 hours. which one is faster. HEEEEEEELLLLLLLLLLLLLL
pantera1 [17]

Answer:

Brook is faster.

Step-by-step explanation:

1 hour=60 minutes

2 hours=2*60=120 minutes

-------------

7 miles in 30 minutes vs 25 miles in 120 minutes

7/30=0.233...mile per minute

25/120=0.20833...mile per minute

4 0
3 years ago
Given the equation P2 = A3, what is the orbital period, in years, for the planet Saturn? (Saturn is located 9.5 AU from the sun.
algol [13]

Answer: 29.28 years

Explanation:

From Kepler's third law the square of orbital period of revolving celestial body is proportional to the cube of semi -major axis from the body  it is revolving about.

P² =A³

Where, P is the orbital period in years and A is the semi-major axis in AU (Astronomical units)

It is given that, For Saturn, A = 9.5 AU. We need to find P

⇒P² = (9.5 AU)³

⇒P² = 857.38

⇒P = 29.28 years

Thus, the orbital period of Jupiter is 29.28 years around the Sun.

5 0
4 years ago
6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP
adelina 88 [10]

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = 360^{o} - 324^{o}

     = 036^{o}

Sum of angles in a triangle = 180^{o}

<P + <Q + <S = 180^{o}

048° + 036^{o} + <S = 180^{o}

84^{o} + <S = 180^{o}

<S  = 180^{o} -  84^{o}

    = 96^{o}

<S = 96^{o}

Applying the Sine rule,

\frac{y}{Sin P} = \frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{y}{Sin P} = \frac{z}{Sin S}

\frac{y}{Sin 48^{o} } = \frac{17}{Sin 96^{o} }

\frac{y}{0.74314} = \frac{17}{0.99452}

⇒ y = \frac{12.63338}{0.99452}

       = 12.703

y = 12.70 km

\frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{x}{Sin 36^{o} } = \frac{17}{Sin 96^{o} }

\frac{x}{0.58779} = \frac{17}{0.99452}

⇒ x = \frac{9.992430}{0.99452}

      = 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

6 0
3 years ago
How long will it take for 750 mg of a sample of radium-225, which has a half-life of about 15 days, to decay to 68 mg?
AleksAgata [21]

Answer:

52 Days

Step-by-step explanation:

6 0
3 years ago
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